A) 42
B) 45
C) 48
D) 54
Correct Answer: D
Solution :
The effective atomic number for\[\text{5}.\text{25}\times \text{1}{{0}^{\text{3}}}\text{min}\] (atomic no. of\[\left[ \text{FL}{{\text{T}}^{-\text{2}}} \right]\]) is 54. Because rhodium is in oxidation state. \[\left[ \text{F}{{\text{L}}^{\text{2}}}{{T}^{-\text{2}}} \right]\] \[\left[ \text{F}{{\text{L}}^{-\text{1}}}{{\text{T}}^{\text{2}}} \right]\] \[\left[ {{\text{F}}^{2}}\text{L}{{\text{T}}^{\text{-2}}} \right]\] \[-\text{273}.\text{15}{}^\circ \text{F}\]electrons from 6 ligands \[-\text{453}.\text{15}{}^\circ \text{F}\] \[-\text{459}.\text{67}{}^\circ \text{F}\]
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